y=-1.(1)
x+y+1=0.(2)
∠B,∠C的内角平分线交点M(0,-1)
设B(b,-1),C(c,-1-c)
k(AM)=3,k(BM)=0,k(AB)=3/(1+b),k(AC)=(3-c)/(1+c),k(BC)=c/(b-c)
[k(AC)-k(AM)]/(1+k(AC)*k(AM)]=[k(AM)-k(AB)]/[1+k(AM)*k(AB)]
bc=15b+20c.(1)
[k(AB)-k(BM)]/[1+k(AB)*k(BM)]=[k(BM)-k(BC)]/[1+k(BM)*k(BC)]
bc=2c-3b.(2)
b=5,c=-5
A(-1,-4),B(5,-1),C(-5,4)以下你会了